Question

Let \(f\) be differentiable and let \(f^{\prime}\left(x_{0}\right)=m .\) Find \(f^{\prime}\left(-x_{0}\right)\) if (a) \(f\) is an odd function. (b) \(f\) is an even function.

Short answer

(a) \( f'(-x_0) = m \), (b) \( f'(-x_0) = -m \).

Step-by-step solution

Understand the Given

We know that the derivative of the function at a point is given by \( f'(x_0) = m \). Our task is to find \( f'(-x_0) \) given different properties of the function \( f \).

Analyze Odd Function Property

For an odd function, \( f(-x) = -f(x) \). We need to differentiate both sides of this equation to find \( f'(-x) \):Differentiating both sides with respect to \( x \), we get: \[ \frac{d}{dx}f(-x) = \frac{d}{dx} (-f(x)) \]Using the chain rule:\(-f'(-x) \cdot (-1) = -f'(x) \), so \( f'(-x) = f'(x) \).

Substitute Given Values for Odd Function Case

Substitute \( x_0 \) into the equation for odd functions:\( f'(-x_0) = f'(x_0) = m \).

Analyze Even Function Property

For an even function, \( f(-x) = f(x) \). We differentiate both sides of this with respect to \( x \):Differentiating both sides:\[ \frac{d}{dx}f(-x) = \frac{d}{dx} f(x) \]Use the chain rule:\(-f'(-x) = f'(x) \), so \( f'(-x) = -f'(x) \).

Substitute Given Values for Even Function Case

Substitute \( x_0 \) into the equation for even functions:\( f'(-x_0) = -f'(x_0) = -m \).

Key concepts

Differentiable Functions

Differential calculus is a fundamental concept in mathematics that explores rates of change. At its core, it revolves around the concept of differentiable functions. A function is differentiable at a point if it has a well-defined derivative at that point, meaning you can represent the change of the function with a real number. Put simply, if you can draw a smooth tangent line at any point on the curve of a function, then the function is differentiable at that point.

To determine if a function is differentiable over an interval, it must not have any abrupt 'jumps' or 'kinks'. It should flow smoothly across its domain. The derivative of a function gives us a new function that describes how the original function changes. For example, the slope of a tangent to the function at any particular point gives you the derivative at that point.

Knowing whether a function is differentiable not only tells us about the shape and gradient of the function but also empowers us to make predictions about future behavior. Differentiability is the stepping stone to solving various complex problems in calculus. And, as shown in our exercise, understanding differentiable properties is crucial for solving tasks requiring the differentiation of even and odd functions.

Odd and Even Functions

Odd and even functions have unique symmetry properties that make differentiating them quite interesting. An even function is symmetric with respect to the y-axis. This means that if you flip it over the y-axis, it looks the same. Mathematically, for a function to be even, it must satisfy the condition:

• \( f(-x) = f(x) \).

Take for example the function \( f(x) = x^2 \). If you substitute \( -x \) for \( x \), \( f(-x) = (-x)^2 = x^2 \), which is equal to \( f(x) \), hence even.

On the flip side, an odd function shows rotational symmetry of 180 degrees around the origin. For odd functions, the condition to fulfill this symmetry is:

• \( f(-x) = -f(x) \).

Consider \( f(x) = x^3 \). Here, replacing \( x \) with \( -x \) results in \( f(-x) = (-x)^3 = -x^3 = -f(x) \).

Understanding whether a function is odd or even simplifies the process of finding derivatives because it allows us to apply specific rules. As noted in the provided exercise, these properties informed the solution to finding the derivatives at \( -x_0 \) for both odd and even functions.

Derivative Properties

The derivative is a powerful tool that provides insight into how a function behaves. It tells us about rates of change and shapes, and it's vital for analyzing the behavior of differentiable functions as they increase or decrease.

When evaluating derivatives, several important properties make our work much simpler. A key property is the chain rule, which is crucial when differentiating composite functions. The chain rule expresses that the derivative of a composition of two functions is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. That is:

• \( \frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x) \).

In our exercise, differentiation of odd and even functions involved utilizing the chain rule to maneuver through the expressions and reach the correct derivative.

Knowing the derivative properties helps avoid complex calculations by allowing us to break down difficult expressions into more manageable steps. These properties lead to clear solutions and enable us to easily handle transformations of functions, like reversing and reflecting in the case of odd and even functions. In summary, mastering derivative properties is pivotal in tackling calculus problems efficiently.